5951 to 6000 MCQs for Lab Technician and Technologist Exam Preparation
5000 Plus MCQs for Lab Technician and Technologists are designed to test the knowledge and proficiency of laboratory professionals who work in the field of clinical laboratory science. These questions cover a wide range of topics related to laboratory science, including anatomy, physiology, microbiology, chemistry, and hematology.
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Questions 5951 to 6000
The β-lipoprotein fraction increases in severe:
- a) Diabetes Mellitus
- b) Uremia
- c) Nephritis
- d) Muscular dystrophy
Explanation:
In severe Diabetes Mellitus, lipid metabolism is disturbed, leading to hyperlipoproteinemia. This is characterized by a significant increase in the β-lipoprotein fraction (LDL) in the blood.
Why other options are incorrect:
Why other options are incorrect:
- • Uremia: Primarily involves the accumulation of nitrogenous waste products like urea, not a primary increase in β-lipoproteins.
- • Nephritis: While it may affect protein levels, the specific hallmark for elevated β-lipoprotein is metabolic syndrome/diabetes.
- • Muscular dystrophy: This is a genetic disorder affecting muscle tissue and protein dystrophin, not lipid transport fractions.
∆9 indicates a double bond between carbon atoms of the fatty acids:
- a) 8 and 9
- b) 9 and 10
- c) 9 and 8
- d) 9 and 12
Explanation:
In fatty acid nomenclature, the Delta (∆) symbol indicates the position of the double bond starting from the carboxyl end (carbon-1). Therefore, ∆9 signifies that the double bond begins at carbon 9 and extends to carbon 10.
Why other options are incorrect:
Why other options are incorrect:
- • 8 and 9: This would be denoted as ∆8.
- • 9 and 8: Carbon numbering always progresses away from the carboxyl group (1, 2, 3…), so this sequence is biologically incorrect.
- • 9 and 12: This would imply a double bond spanning across multiple carbons, which does not follow standard IUPAC fatty acid notation.
The number of carbon atoms in decanoic acid present in butter:
- a) 6
- b) 8
- c) 10
- d) 12
Explanation:
The prefix “deca-” in chemical nomenclature is derived from the Greek word for ten. Therefore, decanoic acid (also known as capric acid) is a saturated fatty acid with a 10-carbon chain.
Why other options are incorrect:
Why other options are incorrect:
- • 6: A 6-carbon saturated fatty acid is known as hexanoic acid (caproic acid).
- • 8: An 8-carbon saturated fatty acid is known as octanoic acid (caprylic acid).
- • 12: A 12-carbon saturated fatty acid is known as dodecanoic acid (lauric acid).
Arachidonic acid contains the number of double bonds:
- a) 2
- b) 3
- c) 4
- d) 5
Explanation:
Arachidonic acid is a polyunsaturated omega-6 fatty acid (20:4, n-6). The “4” in its notation indicates that it contains 4 cis-double bonds at positions 5, 8, 11, and 14.
Why other options are incorrect:
Why other options are incorrect:
- • 2: Fatty acids with 2 double bonds are typically classified as linoleic acid derivatives.
- • 3: Fatty acids with 3 double bonds are typically classified as linolenic acid derivatives.
- • 5: Fatty acids with 5 double bonds (e.g., EPA – Eicosapentaenoic acid) represent a different class of omega-3 fatty acids.
The prostaglandins are synthesized from:
- a) Arachidonic acid
- b) Oleic acid
- c) Linoleic acid
- d) Linolenic acid
Explanation:
Prostaglandins are a group of lipid compounds known as eicosanoids. They are synthesized in the body from Arachidonic acid (a 20-carbon polyunsaturated fatty acid) via the cyclooxygenase (COX) pathway.
Why other options are incorrect:
Why other options are incorrect:
- • Oleic acid: A monounsaturated omega-9 fatty acid that is not a precursor for eicosanoid synthesis.
- • Linoleic acid: While it is an essential fatty acid, it serves primarily as the metabolic precursor that the body uses to synthesize arachidonic acid.
- • Linolenic acid: An omega-3 fatty acid, which is a precursor for different types of eicosanoids (like resolvins), but not the primary precursor for prostaglandins.
Cholesterol is a:
- a) Animal sterol
- b) M.F. C27H46O
- c) 5 methyl groups
- d) All of these
Explanation:
Cholesterol is an animal sterol with the molecular formula C₂₇H₄₆O. Its structure contains five methyl groups. Therefore, all the given statements are correct.
Why other options are correct:
Why other options are correct:
- • Animal sterol: Cholesterol is the principal sterol found in animal tissues.
- • M.F. C₂₇H₄₆O: This is the molecular formula of cholesterol.
- • 5 methyl groups: Cholesterol contains five methyl groups in its steroid structure.
Waxes contain higher alcohols named as:
- a) Methyl
- b) Ethyl
- c) Phytyl
- d) Cetyl
Explanation:
Waxes are esters of long-chain fatty acids with long-chain (higher) monohydric alcohols. Cetyl alcohol (a 16-carbon alcohol) is a classic example of a higher alcohol found in many natural waxes, such as spermaceti wax.
Why other options are incorrect:
Why other options are incorrect:
- • Methyl/Ethyl: These are very short-chain alcohols (1 and 2 carbons respectively) and do not form the long-chain structural waxes found in nature.
- • Phytyl: While it is a long-chain alcohol (a diterpene alcohol), it is primarily associated with the side chain of chlorophyll, not the general class of fatty acid-based waxes.
Lieberman-Burchard reaction is performed to detect:
- a) Cholesterol
- b) Glycerol
- c) Fatty acid
- d) Vitamin D
Explanation:
The Lieberman-Burchard reaction is a colorimetric test used specifically to detect cholesterol. In the presence of acetic anhydride and concentrated sulfuric acid, cholesterol undergoes dehydration and oxidation, producing a characteristic green color.
Why other options are incorrect:
Why other options are incorrect:
- • Glycerol: Typically detected using the acrolein test (production of a pungent odor upon heating with KHSO4).
- • Fatty acid: Usually identified through tests like the copper soap test or by determining iodine/saponification numbers.
- • Vitamin D: While it is a sterol derivative, the classic Lieberman-Burchard test is the gold standard for clinical cholesterol quantification.
Lipose present in the stomach cannot hydrolyze fats owing to:
- a) Alkalinity
- b) Acidity
- c) High acidity
- d) Neutrality
Explanation:
While gastric lipase does have some activity, it is severely restricted in the stomach because the highly acidic environment (pH 1.5–2.5) of the gastric juice is far from the optimal pH required for significant fat hydrolysis, and the lack of emulsifying agents like bile salts prevents effective interaction with dietary fats.
Why other options are incorrect:
Why other options are incorrect:
- • Alkalinity/Neutrality: The environment of the stomach is distinctly acidic, not neutral or alkaline.
- • Acidity: While the stomach is acidic, it is specifically the high acidity that inhibits optimal enzymatic function for fat digestion compared to the alkaline environment of the small intestine where bile and pancreatic lipase act.
Fatty acids are oxidized by:
- a) α-oxidation
- b) β-oxidation
- c) ω-oxidation
- d) All of these
Explanation:
Fatty acids can undergo α-oxidation, β-oxidation, and ω-oxidation. β-oxidation is the major pathway for most fatty acids, while α-oxidation and ω-oxidation are used for specific fatty acids or metabolic circumstances. Therefore, all of these are correct.
Why other options are correct:
Why other options are correct:
- • α-oxidation: Oxidizes certain branched-chain fatty acids, such as phytanic acid.
- • β-oxidation: The major pathway for degradation of fatty acids, producing acetyl-CoA, NADH, and FADH₂.
- • ω-oxidation: An alternative pathway that occurs mainly in the endoplasmic reticulum and oxidizes the terminal (ω) carbon.
The fatty acids containing even number and odd number of carbon atoms as well as the unsaturated fatty acids are oxidized by:
- a) α-oxidation
- b) β-oxidation
- c) ω-oxidation
- d) All of these
Explanation:
β-oxidation is the primary metabolic pathway for the breakdown of most fatty acids. It efficiently processes even-chain, odd-chain (which yields a final propionyl-CoA), and unsaturated fatty acids (with the help of auxiliary enzymes like isomerases and reductases).
Why other options are incorrect:
Why other options are incorrect:
- • α-oxidation: Specialized for branched-chain fatty acids like phytanic acid, not for general even/odd/unsaturated chains.
- • ω-oxidation: A minor pathway that converts fatty acids to dicarboxylic acids, usually when β-oxidation is defective.
Long chain fatty acids are first activated to acyl CoA in the:
- a) Cytosol
- b) Mitochondria
- c) Ribosomes
- d) Microsome
Explanation:
Before fatty acids can be oxidized, they must be “activated” by the enzyme acyl-CoA synthetase. This process occurs in the cytosol of the cell, where the fatty acid is converted into fatty acyl-CoA before it can be transported into the mitochondria via the carnitine shuttle.
Why other options are incorrect:
Why other options are incorrect:
- • Mitochondria: While β-oxidation takes place here, the activation of fatty acids occurs in the cytosol prior to entry.
- • Ribosomes: These are the sites of protein synthesis, not lipid metabolism.
- • Microsome: Generally involved in fatty acid elongation or desaturation, not the primary activation step for β-oxidation.
Long chain acyl CoA penetrates mitochondria in the presence of:
- a) Palmitate
- b) Carnitine
- c) Sorbitol
- d) DNP
Explanation:
Long-chain acyl-CoA molecules cannot directly cross the inner mitochondrial membrane. They require the carnitine shuttle system, where the enzyme Carnitine Palmitoyltransferase I (CPT-I) converts acyl-CoA to acyl-carnitine, allowing it to traverse the membrane.
Why other options are incorrect:
Why other options are incorrect:
- • Palmitate: A long-chain fatty acid, not a transport carrier.
- • Sorbitol: A sugar alcohol involved in different metabolic pathways (e.g., polyol pathway).
- • DNP (2,4-Dinitrophenol): A chemical uncoupler of oxidative phosphorylation, not a lipid transport molecule.
Acyl-CoA dehydrogenase converts Acyl CoA to α-β unsaturated acyl-CoA in presence of the coenzyme:
- a) NAD+
- b) NADP+
- c) ATP
- d) FAD
Explanation:
The first step of β-oxidation is catalyzed by the enzyme Acyl-CoA dehydrogenase. This reaction involves the removal of hydrogen atoms to form a double bond between the α and β carbons, and this process requires FAD (Flavin Adenine Dinucleotide) as the electron acceptor, which is reduced to FADH₂.
Why other options are incorrect:
Why other options are incorrect:
- • NAD+/NADP+: These are typically used in reactions involving hydroxyl groups (like the 3rd step of β-oxidation) rather than the formation of C=C double bonds in this context.
- • ATP: ATP is used for energy coupling and activation steps, not as a redox coenzyme for dehydrogenase reactions.
For the activation of long chain fatty acids the enzyme thiokinase requires the cofactor:
- a) Mg++
- b) Ca++
- c) Mn++
- d) K+
Explanation:
The enzyme thiokinase (also known as acyl-CoA synthetase) requires Magnesium ions (Mg++) as an essential cofactor for its catalytic activity. This enzyme catalyzes the reaction of a fatty acid with ATP and Coenzyme A; the Mg++ ions help stabilize the negative charges on the phosphate groups of the ATP molecule, facilitating the reaction.
Why other options are incorrect:
Why other options are incorrect:
- • Ca++: While important for muscle contraction and signaling, it is not the specific cofactor required for the thiokinase activation of fatty acids.
- • Mn++ / K+: These ions serve various roles in cellular metabolism but are not the primary cofactors for this specific activation enzyme.
ω-oxidation takes place by the hydroxylase in microsomes involving:
- a) Cytochrome b
- b) Cytochrome c
- c) Cytochrome P-450
- d) Cytochrome a3
Explanation:
ω-oxidation is a pathway for fatty acid metabolism that occurs in the endoplasmic reticulum (microsomes) of liver and kidney cells. The initial step, which involves the hydroxylation of the terminal methyl group (omega carbon), is catalyzed by a hydroxylase system that requires the presence of Cytochrome P-450, molecular oxygen, and NADPH.
Why other options are incorrect:
Why other options are incorrect:
- • Cytochrome b/c/a3: These cytochromes are primarily components of the mitochondrial electron transport chain (respiratory chain) and are not involved in the microsomal ω-oxidation of fatty acids.
Carboxylation of acetyl—CoA to malonyl — CoA takes place in presence of:
- a) FAD+
- b) Biotin
- c) NAD+
- d) NADP+
Explanation:
The conversion of Acetyl-CoA to Malonyl-CoA is the rate-limiting step in fatty acid synthesis. This reaction is catalyzed by the enzyme Acetyl-CoA Carboxylase, which requires Biotin (Vitamin B7) as a prosthetic group to carry the carboxyl group during the reaction.
Why other options are incorrect:
Why other options are incorrect:
- • FAD+ / NAD+ / NADP+: These are electron carriers (coenzymes) used in redox reactions. Since the carboxylation reaction is a transfer of a carboxyl group rather than an oxidation-reduction reaction, these are not required.
Malonyl-CoA reacts with the central:
- a) —SH group
- b) —NH2 group
- c) —COOH group
- d) —CH2OH group
Explanation:
During fatty acid synthesis, the Fatty Acid Synthase (FAS) complex features an Acyl Carrier Protein (ACP). The malonyl group from Malonyl-CoA is transferred to the central —SH (sulfhydryl) group of the phosphopantetheine moiety of the ACP to initiate the elongation cycle.
Why other options are incorrect:
Why other options are incorrect:
- • —NH2, —COOH, —CH2OH: These functional groups are not the active sites involved in the thioester linkage required for the movement and processing of acyl groups during the fatty acid synthesis cycle.
Fatty acid synthesis takes place in the presence of the coenzyme:
- a) NAD+
- b) Reduced NAD
- c) NADP+
- d) Reduced NADP
Explanation:
Fatty acid synthesis is a reductive process that requires NADPH (Reduced NADP) as the primary electron donor. NADPH provides the reducing power necessary for the conversion of keto groups to hydroxyl groups and the reduction of double bonds during the elongation cycle of the fatty acid chain.
Why other options are incorrect:
Why other options are incorrect:
- • NAD+/Reduced NAD: These are primarily involved in catabolic (oxidative) pathways like glycolysis and the TCA cycle, rather than anabolic (reductive) synthesis.
- • NADP+: This is the oxidized form, which serves as the electron acceptor, not the electron donor required for synthesis.
Fatty acids are activated to acyl CoA by the enzyme thiokinase:
- a) NAD+
- b) NADP+
- c) CoA
- d) FAD+
Explanation:
The enzyme thiokinase (acyl-CoA synthetase) catalyzes the reaction where a fatty acid reacts with Coenzyme A (CoA) and ATP to form fatty acyl-CoA. CoA provides the essential thiol group that forms a high-energy thioester bond with the fatty acid, “activating” it for further metabolism.
Why other options are incorrect:
Why other options are incorrect:
- • NAD+/NADP+/FAD+: These are coenzymes involved in redox (electron transfer) reactions, not in the transfer or activation of acyl groups via thioester bonds.
Phospholipids help the oxidation of:
- a) Glycerol
- b) Fatty acids
- c) Glycerophosphates
- d) None of these
Explanation:
Phospholipids play an important role in lipid metabolism and can facilitate the oxidation of fatty acids. Certain phospholipids participate in mitochondrial membrane structure and function, supporting the processes involved in fatty acid oxidation.
Why other options are incorrect:
Why other options are incorrect:
- • Glycerol: Glycerol is metabolized through glycerol-3-phosphate and glycolytic pathways rather than being the primary substrate referred to here.
- • Glycerophosphates: These are intermediates involved mainly in phospholipid metabolism, not the principal substrate whose oxidation is facilitated.
- • None of these: Fatty acid oxidation is the intended process associated with this function.
The desaturation and chain elongation system of polyunsaturated fatty acids are greatly diminished in the absence of:
- a) Insulin
- b) Glucagon
- c) Epinephrine
- d) Thyroxine
Explanation:
Insulin plays a vital role in lipid metabolism by promoting lipogenesis. In the absence of insulin (such as in uncontrolled diabetes), the activity of the enzymes responsible for desaturation (e.g., Δ6-desaturase) and chain elongation of polyunsaturated fatty acids is significantly reduced, impairing the body’s ability to synthesize long-chain polyunsaturated fatty acids.
Why other options are incorrect:
Why other options are incorrect:
- • Glucagon / Epinephrine: These are primarily catabolic hormones that stimulate lipolysis, not the synthesis or elongation of fatty acids.
- • Thyroxine: While it regulates metabolic rate, it is not the primary hormonal regulator for the specific desaturase and elongase enzyme systems.
Prostaglandins have a common structure based on prostanoic acid which contains carbon atoms:
- a) 12
- b) 16
- c) 18
- d) 20
Explanation:
Prostaglandins are a group of lipids made at sites of tissue damage or infection. They are derived enzymatically from fatty acids and their basic structural skeleton, prostanoic acid, consists of a 20-carbon chain.
Why other options are incorrect:
Why other options are incorrect:
- • 12, 16, 18: These carbon counts do not represent the fundamental prostanoic acid skeleton, which is specifically defined by its 20-carbon structure.
The carbon chains of prostanoic acid are bonded at the middle of the chain by a:
- a) 5-membered ring
- b) 6-membered ring
- c) 8-membered ring
- d) None of these
Explanation:
The structure of prostanoic acid is characterized by a cyclopentane ring (a 5-membered ring) that joins two carbon side chains together at carbons 8 and 12.
Why other options are incorrect:
Why other options are incorrect:
- • 6-membered/8-membered rings: These ring sizes are not part of the basic skeletal structure of prostaglandins.
- • None of these: Incorrect, because the 5-membered cyclopentane ring is the defining structural feature of the prostanoic acid skeleton.
The enzyme systems for lengthening and shortening, and for saturating and desaturating of fatty acids occur in:
- a) Intestine
- b) Muscle
- c) Kidney
- d) Liver
Explanation:
The liver is the primary site for the metabolic processing of fatty acids. It contains the complex enzyme systems required for the elongation, shortening, saturation, and desaturation of fatty acid chains to meet the body’s physiological needs.
Why other options are incorrect:
Why other options are incorrect:
- • Intestine, Muscle, Kidney: While these organs perform various metabolic functions, they do not possess the comprehensive enzymatic machinery for all these specific fatty acid modifications at the same magnitude as the liver.
Which of the following are classified as essential fatty acids?
- a) Arachidonic acid
- b) Oleic acid
- c) Acetic acid
- d) Butyric acid
Explanation:
Arachidonic acid is considered an essential (or conditionally essential) fatty acid. While the body can synthesize some arachidonic acid from linoleic acid, it is required for the synthesis of eicosanoids and must be obtained through the diet if linoleic acid intake is insufficient.
Why other options are incorrect:
Why other options are incorrect:
- • Oleic acid: It is a non-essential monounsaturated fatty acid that the human body can synthesize.
- • Acetic acid: A short-chain carboxylic acid, not a fatty acid required in the diet.
- • Butyric acid: A short-chain saturated fatty acid that the body can produce and is not considered “essential.”
Prostaglandins are synthesized in the body from:
- a) Myristic acid
- b) Arachidonic acid
- c) Stearic acid
- d) Lignoceric acid
Explanation:
Arachidonic acid (a polyunsaturated omega-6 fatty acid) serves as the primary precursor for the synthesis of prostaglandins, thromboxanes, and leukotrienes (collectively known as eicosanoids) via the cyclooxygenase (COX) pathway.
Why other options are incorrect:
Why other options are incorrect:
- • Myristic acid, Stearic acid: These are saturated fatty acids and do not serve as precursors for prostaglandin synthesis.
- • Lignoceric acid: A very-long-chain saturated fatty acid found in sphingolipids, not involved in eicosanoid synthesis.
All the following saturated fatty acids are present in butter except:
- a) Butyric acid
- b) Caprylic acid
- c) Caproic acid
- d) Capric acid
Explanation:
Butter contains several short- and medium-chain saturated fatty acids, including butyric acid (C4:0), caproic acid (C6:0), and caprylic acid (C8:0). Capric acid (C10:0) is present in milk fat but is not typically listed among the characteristic short-chain fatty acids of butter in basic biochemistry questions. Therefore, the expected answer is Capric acid.
Why other options are incorrect:
Why other options are incorrect:
- • Butyric acid: A characteristic short-chain fatty acid found in butter.
- • Caprylic acid: A saturated C8 fatty acid found in butterfat.
- • Caproic acid: A saturated C6 fatty acid found in butterfat.
Saponification number is:
- a) mg of KOH required to saponify one gm of fat or oil
- b) mg of KOH required to neutralize free fatty acids of one gms of fat
- c) mg of KOH required to neutralize the acetic acid obtained by saponification of one gm of fat after it has been acetylated
- d) None of these
Explanation:
The saponification number (or saponification value) represents the number of milligrams of potassium hydroxide (KOH) required to completely saponify one gram of fat or oil. It is a measure of the average molecular weight of the fatty acids present in the fat.
Why other options are incorrect:
Why other options are incorrect:
- • Acid Value: This refers to the KOH required to neutralize free fatty acids (Option b).
- • Acetyl Value: This refers to the amount of KOH required to neutralize acetic acid released after the acetylation of fats (Option c).
Lipids have the following properties:
- a) Insoluble in water and soluble in fat solvent
- b) High energy content
- c) Structural component of cell membrane
- d) All of these
Explanation:
Lipids are generally insoluble in water but soluble in non-polar organic solvents. They provide a high energy yield and are important structural components of cell membranes. Therefore, all the given statements are correct.
Why other options are correct:
Why other options are correct:
- • Insoluble in water and soluble in fat solvent: This is a characteristic property of most lipids.
- • High energy content: Lipids provide approximately 9 kcal of energy per gram.
- • Structural component of cell membrane: Phospholipids, cholesterol, and glycolipids are important components of biological membranes.
Carbohydrate moiety in cerebrosides is:
- a) Glucose
- b) Sucrose
- c) Galactose
- d) Maltose
Explanation:
Cerebrosides are glycosphingolipids consisting of a ceramide backbone with a single sugar residue. In most cases, this sugar residue is Galactose (often called galactocerebrosides), which is particularly abundant in the myelin sheaths of nerve cells.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: While glucocerebrosides exist (found primarily in non-neuronal tissues), Galactose is the classic and most defining sugar moiety associated with the name “cerebroside.”
- • Sucrose & Maltose: These are disaccharides and are not the sugar components found in simple cerebrosides.
Which of the following is not an unsaturated fatty acid?
- a) Oleic acid
- b) Stearic acid
- c) Linaoleic acid
- d) Palmitic acid
Explanation:
Stearic acid is a saturated fatty acid containing no carbon-carbon double bonds. Oleic acid and linoleic acid are unsaturated fatty acids.
Why other options are incorrect:
Why other options are incorrect:
- • Oleic acid: A monounsaturated fatty acid with one double bond.
- • Linoleic acid: A polyunsaturated fatty acid with two double bonds.
- • Palmitic acid: Also a saturated fatty acid, but the standard single-best answer to this MCQ is stearic acid.
All the following are functions of prostaglandins except
- a) Lowering of B.P
- b) Introduction of labour
- c) Anti inflammatory
- d) Prevention of myocardial infraction
Explanation:
Prostaglandins have several important physiological functions, including regulation of blood pressure and stimulation of uterine contractions during labour. However, prostaglandins are generally associated with promoting inflammation rather than having a general anti-inflammatory action.
Why other options are incorrect:
Why other options are incorrect:
- • Lowering of B.P: Certain prostaglandins, such as PGE₂ and PGI₂, can cause vasodilation and lower blood pressure.
- • Introduction of labour: Prostaglandins are used clinically to promote cervical ripening and induce labour.
- • Prevention of myocardial infraction: Prostaglandin-mediated vasodilation and inhibition of platelet aggregation, particularly by prostacyclin (PGI₂), can have protective cardiovascular effects.
Calorific value of lipids per gm is:
- a) 4 Kcal
- b) 8 Kcal
- c) 9 Kcal
- d) None of these
Explanation:
Lipids are the most energy-dense macronutrients. When oxidized in the body, one gram of fat yields approximately 9 Kcal (kilocalories) of energy, which is more than double the energy provided by carbohydrates or proteins (both yield approximately 4 Kcal/g).
Why other options are incorrect:
Why other options are incorrect:
- • 4 Kcal: This is the energy density typical of carbohydrates and proteins.
- • 8 Kcal: This is an incorrect approximation; the standard biological energy value for fats is 9 Kcal/g.
All the following are ketones (ketoses) except:
- a) Xylulose
- b) Ribolose
- c) Erythrose
- d) Fructose
Explanation:
Sugars are classified based on their functional group. Ketoses contain a ketone group, while aldoses contain an aldehyde group. Erythrose is a tetrose sugar with an aldehyde group, making it an aldose, not a ketose.
Why other options are incorrect:
Why other options are incorrect:
- • Xylulose, Ribulose, & Fructose: These are all classic examples of ketose sugars containing a ketone functional group.
Saponification is:
- a) Hydrolysis of fats by alkali
- b) Hydrolysis of glycerol by lipases
- c) Esterification
- d) Reduction
Explanation:
Saponification is the chemical reaction where fats or oils (triglycerides) are reacted with an alkali (like NaOH or KOH) to produce glycerol and the salts of fatty acids, commonly known as soap.
Why other options are incorrect:
Why other options are incorrect:
- • Hydrolysis by lipases: This is an enzymatic breakdown process, not chemical saponification.
- • Esterification: This is the reverse process, where an acid and an alcohol react to form an ester.
- • Reduction: This involves the gain of electrons or hydrogen, which is not the mechanism of saponification.
Number of ml of 0.1 N KOH required to neutralize fatty acids from 5 gms of fat:
- a) Iodine number
- b) Polenske number
- c) Reichert-Miessl number
- d) None of these
Explanation:
The amount of 0.1 N KOH required to neutralize the free fatty acids in a specified amount of fat is not measured by the iodine number, Polenske number, or Reichert-Miessl number. Therefore, the correct answer is None of these.
Why other options are incorrect:
Why other options are incorrect:
- • Iodine number: Measures the grams of iodine absorbed by 100 g of fat and indicates the degree of unsaturation.
- • Polenske number: Measures the milliliters of 0.1 N alkali required to neutralize the water-insoluble volatile fatty acids obtained from a specified amount of fat.
- • Reichert-Miessl number: Measures the milliliters of 0.1 N alkali required to neutralize the water-soluble volatile fatty acids obtained from 5 g of fat.
Hydrated density of HDL lipoproteins is
- a) 0.94 gm/ml
- b) 0.94–1.006 gm/ml
- c) 1.006–1.063 gm/ml
- d) 1.063–1.21 gm/ml
Explanation:
HDL (High-Density Lipoprotein) has the highest density among the major plasma lipoproteins. Its hydrated density ranges from approximately 1.063 to 1.21 g/mL.
Why other options are incorrect:
Why other options are incorrect:
- • 0.94 gm/ml: This value is below the density range of HDL.
- • 0.94–1.006 gm/ml: This range corresponds mainly to chylomicrons and VLDL.
- • 1.006–1.063 gm/ml: This is the approximate density range of LDL.
Saponification number indicates:
- a) Unsaturation in fat
- b) Average M.W of fatty acid
- c) Acetyl number
- d) Acid number
Explanation:
The saponification number is a measure of the average molecular weight (chain length) of the fatty acids present in a fat or oil. Because the number of ester bonds per gram of fat increases as the chain length of fatty acids decreases, a higher saponification number indicates that a fat contains a higher proportion of short-chain fatty acids (lower average molecular weight).
Why other options are incorrect:
Why other options are incorrect:
- • Unsaturation: This is measured by the Iodine number.
- • Acetyl number: Measures the number of hydroxyl groups in fats.
- • Acid number: Measures the amount of free fatty acids present in a sample.
Acrolein Test is positive for:
- a) Glycerol
- b) Prostaglandins
- c) Carbohydrates
- d) Proteins
Explanation:
The Acrolein test is a classic qualitative test for the presence of glycerol or fats. When glycerol is heated with a dehydrating agent (like potassium bisulfate), it undergoes dehydration to form acrolein (acrylic aldehyde), which has a characteristic burnt, pungent odor.
Why other options are incorrect:
Why other options are incorrect:
- • Prostaglandins, Carbohydrates, & Proteins: These molecules do not contain the glycerol backbone or undergo the specific dehydration reaction to form acrolein.
Iodine number denotes:
- a) Degree of unsaturation
- b) Saponification number
- c) Acid number
- d) Acetyl number
Explanation:
The iodine number (or iodine value) is the mass of iodine in grams that is consumed by 100 grams of a chemical substance. It is a measure of the degree of unsaturation in a fat or oil, as iodine reacts with the double bonds present in unsaturated fatty acid chains.
Why other options are incorrect:
Why other options are incorrect:
- • Saponification number: Measures average molecular weight.
- • Acid number: Measures free fatty acids.
- • Acetyl number: Measures hydroxyl groups.
Which of the following provides the maximum energy when metabolized?
- a) Fats
- b) Carbohydrates
- c) Proteins
- d) Nucleic acids
Explanation:
Fats (lipids) provide the highest energy density, yielding approximately 9 kcal per gram, whereas carbohydrates and proteins yield approximately 4 kcal per gram.
Why other options are incorrect:
Why other options are incorrect:
- • Carbohydrates: These are the body’s primary and preferred energy source, but they contain less than half the energy per gram compared to fats.
- • Proteins: These are primarily used for building and repairing tissues; they provide the same energy per gram as carbohydrates.
- • Nucleic acids: DNA and RNA are used for genetic storage and protein synthesis, not as a primary source of fuel for the body.
Lecithins are composed of:
- a) Glycerol + Fatty acids + Phosphoric acid + Choline
- b) Glycerol + Fatty acids + Phosphoric acid + Ethanolamine
- c) Glycerol + Fatty acids + Phosphoric acid + Serine
- d) Glycerol + Fatty acids + Phosphoric acid + Betaine
Explanation:
Lecithins, also known as phosphatidylcholines, are a type of phospholipid consisting of a glycerol backbone attached to two fatty acids, a phosphate group, and a choline molecule.
Why other options are incorrect:
Why other options are incorrect:
- • Ethanolamine: This combination forms Phosphatidylethanolamine (Cephalin), not Lecithin.
- • Serine: This combination forms Phosphatidylserine.
- • Betaine: While related to choline metabolism, it is not a structural component of the lecithin molecule.
Sphingomyelins are composed of fatty acids, phosphoric acid and:
- a) Sphingosine and choline
- b) Glycerol and sphingosine
- c) Glycerol and Serine
- d) Glycerol and Choline
Explanation:
Sphingomyelins are a type of sphingophospholipid. Unlike glycerophospholipids, they do not contain glycerol; instead, they contain the amino alcohol sphingosine. The structure consists of sphingosine, a fatty acid, phosphoric acid, and choline.
Why other options are incorrect:
Why other options are incorrect:
- • Glycerol options: Sphingomyelins are characterized by the absence of glycerol; they use sphingosine as the backbone.
Depot fats of mammalian cells comprise mostly of:
- a) Cholesterol
- b) Cholesterol esters
- c) Triacyl glycerol
- d) Phospholipids
Explanation:
Triacylglycerols (also known as triglycerides) are the main storage form of fats in mammalian adipose tissue, serving as the primary fuel depot for the body.
Why other options are incorrect:
Why other options are incorrect:
- • Cholesterol & Cholesterol esters: These are primarily structural components of cell membranes or precursors for steroid hormones, not energy storage molecules.
- • Phospholipids: These are the essential structural components of all cellular membranes, not energy reserves.
When choline of lecithin is replaced by ethanolamine, the product is:
- a) Sphingomyelin
- b) Cephalin
- c) Plasmalogens
- d) Lysolecithin
Explanation:
Cephalin (also known as phosphatidylethanolamine) is structurally similar to lecithin (phosphatidylcholine), but it contains ethanolamine instead of the choline group found in lecithin.
Why other options are incorrect:
Why other options are incorrect:
- • Sphingomyelin: Contains a sphingosine backbone, not glycerol.
- • Plasmalogens: Characterized by an ether linkage at the sn-1 position of the glycerol backbone.
- • Lysolecithin: This is formed by the removal of one fatty acid chain from lecithin (by phospholipase A2), not by swapping the head group.
Which of the following is a hydroxy fatty acid?
- a) Oleic acid
- b) Ricinoleic acid
- c) Caproic acid
- d) Stearic acid
Explanation:
Ricinoleic acid is a fatty acid that contains a hydroxyl (-OH) group on its carbon chain (specifically at the 12th carbon), making it a hydroxy fatty acid. It is the primary component of castor oil.
Why other options are incorrect:
Why other options are incorrect:
- • Oleic acid: A monounsaturated fatty acid without a hydroxyl group.
- • Caproic acid: A short-chain saturated fatty acid.
- • Stearic acid: A long-chain saturated fatty acid.
Acrolein test is answered by:
- a) Cholesterol
- b) Glycerol
- c) Glycosides
- d) Sphingosine
Explanation:
The Acrolein test is a general test for the presence of glycerol or fats containing glycerol. When glycerol is heated with a dehydrating agent like potassium bisulfate (KHSO4), it undergoes dehydration to form acrolein, which has a distinct, unpleasant, pungent odor.
Why other options are incorrect:
Why other options are incorrect:
- • Cholesterol: Responds to tests like Salkowski’s or Liebermann-Burchard test, not the acrolein test.
- • Glycosides: These contain sugars and do not produce acrolein upon dehydration.
- • Sphingosine: A complex amino alcohol that does not contain the glycerol backbone required to yield acrolein.
The smell of fat turned rancid is due to:
- a) Presence of vit E
- b) Presence of quinones
- c) Phenols
- d) Volatile fatty acids
Explanation:
Rancidity in fats, particularly hydrolytic rancidity, is caused by the release of short-chain, volatile fatty acids (such as butyric or caproic acid) from the triglyceride molecule, often due to hydrolysis by enzymes or moisture. These compounds are responsible for the unpleasant, pungent odor associated with rancid fat.
Why other options are incorrect:
Why other options are incorrect:
- • Vitamin E: Acts as an antioxidant that actually prevents or delays rancidity.
- • Quinones/Phenols: These are chemical compounds not directly associated with the hydrolytic breakdown process that produces the characteristic smell of rancidity.
Phospholipids are important cell membrane components because
- a) They have glycerol
- b) They can form bilayers in water
- c) They have both polar and non polar portions
- d) They combine covalently with proteins
Explanation:
Phospholipids are amphipathic molecules, meaning they contain both hydrophilic (polar) and hydrophobic (non-polar) portions. In water, they spontaneously arrange themselves into bilayers, forming the basic structure of cell membranes.
Why other options are incorrect:
Why other options are incorrect:
- • They have glycerol: Not all phospholipids contain glycerol; some, such as sphingomyelin, contain sphingosine.
- • They have both polar and non polar portions: This is a characteristic of phospholipids and explains their membrane-forming property, but the question specifically emphasizes their ability to form bilayers in water.
- • They combine covalently with proteins: Phospholipids generally interact with membrane proteins through non-covalent interactions rather than forming covalent bonds with them.
Answer Key Summary
The questions are typically designed to assess the technical skills and knowledge required for the laboratory profession, including the ability to analyze laboratory test results, perform laboratory procedures, and maintain laboratory equipment.
To prepare for these MCQs, candidates should have a thorough understanding of the key concepts and principles of laboratory science. They should also be familiar with common laboratory equipment and procedures, as well as laboratory safety protocols.
Candidates may also benefit from studying specific laboratory science textbooks or taking online courses that cover the material tested in the MCQs. Additionally, practicing sample MCQs and reviewing the answers can help candidates identify areas where they may need to improve their knowledge or skills.
Overall, the MCQs for lab technologists are designed to be challenging and comprehensive, requiring candidates to demonstrate a high level of proficiency in the field of laboratory science.
📚 This article may have been prepared using trusted medical literature, clinical guidelines, textbooks, and peer-reviewed sources.
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